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Let $A^A$ be the set of all functions from $A$ to $A$. Prove $\mathcal{P}(A)$ and $A^A$ have the same cardinality for infinite sets $A$. ($\mathcal{P}(A)$ is the power set of $A$.)

Is it possible to show this without invoking axiom of choice? It can be proven with $A \times A \sim A$ for infinite sets $A$ but that is equivalent to the axiom of choice.

Qi Zhu
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    Great question! I suspect this does indeed require some choice. Towards a choiceless counterexample, note that if $A$ is amorphous then $\mathcal{P}(A)$ is equinumerous to twice the cardinality of the set of finite subsets of $A$; but there's no obvious reason for $A^A$ to be that small. For example, $A$ could be amorphous yet admit a partition into $k$-size sets for fixed $k$ $T_i$, together with a unary function $f$ which "cycles" each $T_i$: $f$ is injective and $f$ maps $T_i$ into $T_i$. Then there are at least $k$-times as many functions from $A$ to $A$ as there are finite subsets of $A$. – Noah Schweber Nov 15 '17 at 16:27
  • @Noah: Yeah, so great it was asked before. Several times. – Asaf Karagila Nov 15 '17 at 16:29
  • @AsafKaragila I'm not surprised, but I didn't find a duplicate ... – Noah Schweber Nov 15 '17 at 16:31
  • @Noah: I already have two. (Edit: the second one was a dud.) – Asaf Karagila Nov 15 '17 at 16:32
  • @AsafKaragila My point was more that the OP shouldn't be faulted too much for not finding them if I can't find them. – Noah Schweber Nov 15 '17 at 16:35
  • Oh, sure, with that I agree. – Asaf Karagila Nov 15 '17 at 16:35

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